Question

A cube is placed inside an electric field, $\vec{E}=150y^2\hat{j}$. The side of the cube is $0.5m$ and is placed in the field as shown in the given figure. The charge inside the cube is:

• A. $3.8\times 10^{-11}C$
• B. $8.3\times 10^{-11}C$
• C. $3.8\times 10^{-12}C$
• D. $8.3\times 10^{-12}C$

Answer 1

Answer: (B)

As electric field is in y-direction so electric flux is only due to top and bottom surface
Bottom surface $y=0$
$\Rightarrow E=0$
$\Rightarrow \varphi =0$
Top surface $y=0.5m$
$\Rightarrow E=150(.5{)}^2=\frac{150}{4}$
Now flux $\varphi =EA=\frac{150}{4}(.5{)}^2=\frac{150}{16}$
By Gauss's law $\varphi =\frac{Q_{\text{in }}}{{ϵ}_0}$
$\frac{150}{16}=\frac{Q_{\text{in }}}{{ϵ}_0}$
$Q_{\text{in }}=\frac{150}{16}\times 8.85\times 10^{-12}$
$=8.3\times 10^{-11}C$